\newproblem{lay:5_3_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	$A$ is a $3\times 3$ matrix with two eigenvalues. Each eigenspace is one-dimensional. Is $A$ diagonalizable? Why?
}{
  % Solution
	According to Theorem 5.3.7b, a matrix is diagonalizable if and only if the sum of the dimensions of all the eigenspaces is equal to the number of rows and columns of the matrix $A$.
	Since in this case $1+1=2\ne 3$, $A$ is not diagonalizable.
}
\useproblem{lay:5_3_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
